The equation to the normal to the curve y = sin x at (0, 0) is

Question

The equation to the normal to the curve y = sin x at (0, 0) is (A) x = 0 (B) y = 0 (C) x + y = 0 (D) x - y = 0.

Tardigrade Admin · Accepted Answer

y = sin x ⇒ (dy/dx) = cos x At (0,0), (dy/dx) = cos 0=1 ∴ slope of the normal = (1/-1) = -1 ∴ normal at (0, 0) is y - 0 = - 1(x - 0) = - x i.e., x + y = 0

Q. The equation to the normal to the curve $y = sin x$ at (0, 0) is

$x = 0$

$y = 0$

$x + y = 0$

$x - y = 0.$

$y = s in x$
$\Rightarrow \frac{d y}{d x} = cos x$
At $(0, 0), \frac{d y}{d x} = cos 0 = 1$
$∴$ slope of the normal $= \frac{1}{- 1} = - 1$
$∴$ normal at $(0, 0)$ is
$y - 0 = - 1 (x - 0) = - x$ i.e., $x + y = 0$

Q. The equation to the normal to the curve y=sinx at (0, 0) is

x=0

y=0

x+y=0

x−y=0.